Talk:Triangular number

Popular Culture

Personally, I wouldn't consider the bible being "pop culture" — Preceding unsigned comment added by 211.29.185.4 (talk) 06:08, 15 June 2007 (UTC)

Yeah, it’s religious, and been around for hundreds of years. So saying that it’s popular culture doesn’t make sense. — Preceding unsigned comment added by 2601:140:8a00:2059:fde6:9b72:9a53:1d4 (talk) 13:01, 1 March 2019 (UTC)

An elegant property

Not sure if this is worth putting in, but a nice fact. Form a sequence of numbers, starting from one, then repeatedly multiply by nine and add one to get the next term. All the numbers in this sequence are triangular. —Preceding unsigned comment added by 82.6.96.22 (talk) 18:00, 3 May 2011 (UTC)

I found a proof! The result is always 3n+1 th triangular number! 115.250.59.221 (talk) 11:50, 5 January 2013 (UTC)

Might be worth mentioning in the article, but only if we can find a reliable source for it. —David Eppstein (talk) 17:04, 5 January 2013 (UTC)

Did not have time to publish it then, but now here it goes: [looks my IP has changed]

Basic formula = (n^2 + n)/ 2

Now what we want to say is that: 9(n^2 + n/ 2) + 1 = 9n^2 + 9n + 2/ 2 is also in the form of n^2 + n/ 2. Lets see:

9n^2 + 9n + 2/ 2 = 9n^2 + 6n + 1 + 3n + 1/ 2 = (9n^2 + 6n + 1) + (3n + 1)/ 2 = (3n +1)^2 + (3n + 1)/ 2

Wow! if we plug in 3n + 1 in the formula we get the result! For example 1st triangular no. = 1 therefore (3*1)+1 = 4th triangular no. must be (9*1)+1 = 10, which it is! 117.227.200.134 (talk) 17:45, 13 January 2013 (UTC)

Fine, but we need a reliable source before we can add it to the article; otherwise, it is original research. —David Eppstein (talk) 18:40, 13 January 2013 (UTC)

What the hell? We need to add informative content and blah blah, for which it is qualified. Secondly, we should add information which is provably true or has some sources to rely upon, which it also qualifies. Why do we need to hesitate to add it, when it comfortably with the basic rules. And still if you are too strict for rules, there exists a godfather rule called, WP:Ignore all rules. 117.226.193.247 (talk) 14:02, 14 January 2013 (UTC)

It seems likely to me that in this case the sources really do exist. So you just need to find them. —David Eppstein (talk) 15:51, 14 January 2013 (UTC)

Found them: [1] [2] ! The rest lies with you as I don't know LATEX, hope you do it, I am going to bed. 117.226.193.247 (talk) 16:35, 14 January 2013 (UTC)

Flipping through the feedback, I really consider one reader's demand of the first 100th tri. no. in the article, much like it is in prime number and other such articles. [And I think the article needs expansion with more helpful illustrations.] What do you think? 117.226.193.247 (talk) 16:39, 14 January 2013 (UTC)

These numbers are the repunits in base 9; see (sequence A002452 in the OEIS), which notes that they are all triangular numbers. —David Eppstein (talk) 16:36, 15 January 2013 (UTC)

An elegant property #5

n is equal to ${\frac {{\sqrt {8x+1}}-1}{2}}$, such that T_n = x! Let's see how:

First, the basic formula:

n(n+1) / 2 = (n² + n) / 2 = x

= 8x = 4(n²+n) = 4n²+4n [Multiplying by 8 on both sides]

= 8x + 1 = 4n² + 4n + 1 [Adding 1 to both sides]

= 8x + 1 = 4n² + 2n + 2n + 1

= 8x + 1 = 2n(2n + 1) + 1(2n + 1)

= 8x + 1 = (2n + 1)(2n + 1) = (2n + 1)²

= ${\sqrt {8x+1}}=2n+1$ [Square root of both sides]

= ${\sqrt {8x+1}}-1=2n$ [Subtracting 1 to both sides]

= ${\frac {{\sqrt {8x+1}}-1}{2}}=n$ [Dividing 2 on both sides]

Bingo! Let try it with an example, where x = 10:

$n={\frac {{\sqrt {8x+1}}-1}{2}}={\frac {{\sqrt {80+1}}-1}{2}}={\frac {9-1}{2}}=4$

Yoho! The 4th tri. no. is 10, as we found out! 117.227.149.197 (talk) 14:06, 28 January 2013 (UTC)

Yes, this is just what you get if you apply the quadratic formula to the formula for triangular numbers. Hut 8.5 18:37, 28 January 2013 (UTC)

Alternate notation

Does anyone else support the inclusion of other, perhaps less common, forms of notation? I was able to dig up discussion here as well as some additional support for the n? notation here. 67.171.222.203 (talk) 22:53, 15 April 2014 (UTC)

Exponential functions of Triangular numbers

These formulas apply to any x > 0

T(x²) = T(x-1)² + T(x)²

T(1) = 0² + 1² = 1
T(4) = 1² + 3² = 10
T(9) = 3² + 6² = 45

PROOF
T(x²) = [x² * (x² + 1)] / 2 = [x⁴ + x²] / 2

T(x-1)^2 = [x⁴ - 2x³ + x²] / 4
T(x)^2 = [x⁴ + 2x³ + x²] / 4

T(x)² + T(x-1)²
= [2x⁴ + 2x²] / 4
= [x⁴ + x²] / 2

edit: Whoops, just noticed that this one is redundant. My bad.

Next, any T(x)² is equal to the sum of cubes from 1 to x
T(1)² = 1³ = 1 (1²)
T(2)² = 1³ + 2³ = 9 (3²)
T(3)² = 1³ + 2³ + 3³ = 1 + 8 + 27 = 36 (6²)
1 + 8 + 27 + 64 = 100 (10²)

I have not found a formula to express this (as is expected with iterative functions) but it can be verified in about 20 lines of java code.

Finally, 6*T(x) + 1 = (x+1)³ - x³
2³ - 1³ = 8 - 1 = 7 = 6 * T(1) + 1
3³ - 2³ = 27 - 8 = 19 = 6 * T(2) + 1

PROOF
(x+1)³ = x³ + 3x² + 3x + 1
-x³ = 3x² + 3x + 1

6 * T(x) + 1 = 6 * (x² + x) / 2 + 1 = 3x² + 3x + 1

Hopefully these can be put to use. They seem to be similar in structure to existing algebraic formulas, but a bit more straightforward. Jozbornn (talk) 06:50, 2 March 2015 (UTC)

The sum of cubes fact that you state is well known — see squared triangular number. But if you want any of this to be incorporated into the article (and why else would you be posting it here?) you'll need to find reliable sources that already state these facts — otherwise it would be original research. —David Eppstein (talk) 07:24, 2 March 2015 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Triangular number/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Comment(s)

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Currently one long unstructured section consisting of a sequence of unrelated facts. Needs organization and editing. —David Eppstein 01:44, 24 April 2007 (UTC) Add Fermat's Theorem on triangular numbers (first proved by Gauss). I think that a mention of the theorem that every positive integer can be represented as the sum of at most 3 triangular numbers, stated by Fermat and first proved by Gauss in Disquisitiones Arithmeticae, should be added to this page. Perhaps also a reference to its generalization, also stated by Fermat and proved by Cauchy (?) after work by others that every positive integer was a sum of at most 3 triangular numbers, 4 square numbers, 5 pentagonal numbers, etc. Yes I know the history of the 4-square theorem, but I don't think that would be appropriate in the page on triangular numbers. Achava 00:44, 3 May 2007 (UTC) I would consider myself an expert on this topic, and just wanted to note that think this article is extremely well written. I would love to see its author(s) contribute to more content like it. — Preceding unsigned comment added by 24.19.214.183 (talk) 20:26, 10 February 2013 (UTC)

Last edited at 20:27, 10 February 2013 (UTC). Substituted at 02:39, 5 May 2016 (UTC)

Easy formula

Why isn't there a easy formula? Like f(x)=(x/2)(x+1) works very well. 178.121.66.129 (talk) 17:39, 11 March 2019 (UTC)

That's the second-to-last of the terms in the big formula at the start of the "Formula" section. It's also repeated a couple more times in the same section. —David Eppstein (talk) 17:56, 11 March 2019 (UTC)

It is a little confusing, because it looks like just one big equation, rather than three different ways to write the same thing. Sometimes we put them all on separate lines, but perhaps its just done this way to save space since two of the three identities are just one term. —Soap— 03:45, 7 April 2020 (UTC)

Easier proof

Is there a reason why there isn't an easier proof on the page? I mean, it can be solved by anyone who saw $\sum _{x=1}^{n}x=n(n+1)/2$ in school; by induction.

Extended content

Obviously 1³ = 1². if applies for $n$, then it applies for $(n+1)$ as well; so $(\sum _{x=1}^{n+1}x)^{2}=\sum _{x=1}^{n+1}x^{3}$ seperate $n+1$ $((\sum _{x=1}^{n}x)+(n+1))^{2}=(\sum _{x=1}^{n}x^{3})+(n+1)^{3}$ considering we assume it applies for $n$, we can substitute $\sum _{x=1}^{n}x^{3}=(\sum _{x=1}^{n}x)^{2}$ $((\sum _{x=1}^{n}x)+(n+1))^{2}=(\sum _{x=1}^{n}x)^{2}+(n+1)^{3}$ solving the quadratic equation $(\sum _{x=1}^{n}x)^{2}+2(\sum _{x=1}^{n}x)(n+1)+(n+1)^{2}=(\sum _{x=1}^{n}x)^{2}+(n+1)^{3}$ remove $(\sum _{x=1}^{n}x)^{2}$ from both sides $2(\sum _{x=1}^{n}x)(n+1)+(n+1)^{2}=(n+1)^{3}$ considering $(\sum _{x=1}^{n}x)=n(n+1)/2$ $2n(n+1)(n+1)/2+(n+1)^{2}=(n+1)^{3}$ as $2*a*b*b/2=a*b^{2}$ $n(n+1)^{2}+(n+1)^{2}=(n+1)^{3}$ next, we single out $(n+1)^{2}$ as $a*b^{2}+b^{2}=b^{2}(a+1)$ $(n+1)^{2}(n+1)=(n+1)^{3}$ which proves the theorem

Qube0 (talk) 17:18, 5 January 2020 (UTC)

@Qube0: Easier than what? Other than a graphical illustration, there's no proof in this article at all, which is perfectly reasonable. Inductive proofs of identities like this aren't generally very enlightening and just tend to clutter up the article. On a side note, your proof is full of errors – most seriously, starting by writing what you're trying to ultimately prove, rather than winding up with that as the final result. –Deacon Vorbis (carbon • videos) 17:44, 5 January 2020 (UTC)

non-integer triangular numbers

Since the ${\frac {n(n+1)}{2}}$ formula can be evaluated for all real numbers, both positive and negative, integer or non-integer, is there any special term for the numbers that lie along the resulting graph? x^2 evaluates to the same value regardless of sign, so the resulting curve is symmetrical across the axis of x=-1/2. But Im more interested in the fractional terms and if there is anything special about them. e.g. the π^th "triangular number" is approximately 6.50559852734. If there's anything to this, it certainly would be worth mentioning, but I couldnt find anything about "generalized triangular numbers" anywhere. Thank you, —Soap— 03:42, 7 April 2020 (UTC)

Proposed merge from Termial

Termials and triangular numbers are the same thing. Since Wikipedia articles are about concepts not names (WP:NOTDICT), they should have together only one article. Therefore, I am proposing a merge. I think that termial is also a rather obscure term for what is much more widely known as a triangular number, so the merge should go from termial to triangular number. Any other opinions? —David Eppstein (talk) 19:26, 20 August 2020 (UTC)

Support I wouldn't say they are the same thing. One is a sequence while the other a series, and termial can apparently apply to non-integers. However, they are certainly intertwined, and Termial might be better as a section of the Triangular Number article. The term "termial" doesn't seem to appear in much mathematical literature though, so I question its notability in the first place. Kstern (talk) 20:54, 3 September 2020 (UTC)

Support They're the same thing! If someone wants to know more about termials, and they read the article about triangular numbers, that will serve their purpose just fine. — Preceding unsigned comment added by 73.241.189.0 (talk) 17:00, 24 September 2020 (UTC)

• Support it doesn't make much difference whether the concept is presented as a sequence or a series, the two are clearly functionally equivalent and this article does already note that triangular numbers can be represented as a series. Triangular number is definitely the primary usage and I suggest we just note somewhere that they are sometimes known as termials (The Art of Computer Programming is very influential). Termial doesn't actually specify how it can be calculated for non-integers. Hut 8.5 19:11, 24 September 2020 (UTC)